By Theorems REF , REF , REF , REF , REF , the algebras \(D_2^{t_{\alpha , \beta }}\) , \(D_2^{t_{\gamma ,\gamma }}\) , \(D_2^{t_{\delta ,0}}\) , \(C_2^{t_{\epsilon ,1}}\)  and \(UT_2\)  generate varieties of exponential growth. Hence, if \(c_n^{tr}(A)\)  is polynomially bounded, then \( D_2^{t_{\alpha , \beta }}\) , \(D_2^{t_{\gamma ,\gamma }}\) , \( D_2^{t_{\delta ,0}}\) , \(C_2^{t_{\epsilon ,1}}, UT_2 \notin {tr}(A)\) , for any \(\alpha \) , \(\beta \) , \(\gamma \) , \(\delta \in F \setminus \lbrace  0 \rbrace \) , \(\alpha \ne \beta \) , \(\epsilon \in F\) .
Conversely suppose that \( D_2^{t_{\alpha , \beta }}\) , \(D_2^{t_{\gamma ,\gamma }}\) , \( D_2^{t_{\delta , 0}}\) , \(C_2^{t_{\epsilon ,1}}, UT_2 \notin {tr}(A)\) , for any \(\alpha \) , \(\beta \) , \(\gamma \) , \( \delta \in F \setminus \lbrace  0 \rbrace \) , \(\alpha \ne \beta \) , \(\epsilon \in F\) . Since we are dealing with codimensions, and these do not change under extensions of the base field, we may assume that the field \(F\)  is algebraically closed. By Theorem REF , we get that
\(A = M_{n_1}(F) \oplus \cdots \oplus M_{n_k}(F) + J, \ \ k \ge 1,\) 
and there exist constants \(\alpha _i\)  such that, for \(a_i \in M_{n_i}(F)\) , we have
\(tr(a_1, \ldots , a_k) = \sum _{i = 1}^{k} t_{\alpha _i}(a_i).\) 
Since \(D_2^{t_{\gamma ,\gamma }} \notin {tr}(A)\) , for any \(\gamma \in F \setminus \lbrace  0 \rbrace \) , and since, by Lemma REF , we have that, for \(n\ge 2\) , \(D_2^{t_{\gamma ,\gamma }} \in {tr}(M^{t_\gamma }_n) \subseteq {tr}(A) \) , we get that \(n_i = 1\) , for every \(i = 1\) , ..., \(k\) . Hence
\(A=A_1\oplus \cdots \oplus A_k + J\) 
where for every \(i=1\) , ..., \(k\) , \(A_i\cong F\)  and the trace on it is \(t_{\alpha _i}\) .
Since, for any \(\alpha \in F\) , \(C_2^{t_{\alpha ,1}} \notin {tr}(A)\) , by Lemma REF  we must have that the trace vanishes on \(J\) .
Now, if for any \(i = 1\) , ..., \(k\) , the trace on \(A_i\)  is zero, since \(UT_2 \notin {tr}(A)\) , then, for any \(i \ne j\) , we must have \(A_i J A_j = 0\) . Hence, for \(n \ge 1\) , \(c_n^{tr}(A) = c_n(A)\)  is polynomially bounded (see, for instance [1]}) and we are done in this case.
Hence, we may assume that there exists \(i\)  such that the trace on \(A_i\)  is \(t_{\alpha _i}\) , with \(\alpha _i \ne 0\) .
Let \(F_\alpha \)  denote the field \(F\)  endowed with the trace \(t_\alpha \) . We claim that \(F_\alpha \oplus F_\beta \)  is isomorphic to \(D_2^{t_{\alpha , \beta }}\)  if \(\alpha \ne \beta \)  (notice that \(\beta \)  could be zero) and to \(D_2^{t_{\alpha , \alpha }}\)  otherwise. Here we shall denote by \(t\)  the trace map on \(F_{\alpha } \oplus F_{\beta } \)  defined as \(t((a,b)) = t_{\alpha }(a) + t_{\beta }(b)\) , for all \((a,b) \in F_\alpha \oplus F_\beta \) . In order to prove the claim, let us consider the linear map \(\varphi \colon D_2 \rightarrow F_\alpha \oplus F_\beta \)  such that
\(\varphi \begin{pmatrix} 1 & 0 \\ 0 &0 \end{pmatrix} = (1,0) \ \ \ \ \ \ \ \ \ \mbox{and} \ \ \ \ \ \ \ \ \ \varphi \begin{pmatrix} 0 & 0 \\ 0 &1 \end{pmatrix} = (0,1).\) 
It is easily seen that \(\varphi \)  is an isomorphism of algebras.
Now, if \(\alpha \ne \beta \) , we have that
\(\varphi \left( t_{\alpha , \beta } \begin{pmatrix} 1 & 0 \\ 0 &0 \end{pmatrix} \right) = \varphi (\alpha ) = (\alpha , \alpha ) = t(1,0) = t \left( \varphi \begin{pmatrix} 1 & 0 \\ 0 &0 \end{pmatrix} \right),\) 
\(\varphi \left( t_{\alpha , \beta } \begin{pmatrix} 0 & 0 \\ 0 &1 \end{pmatrix} \right) = \varphi (\beta ) = (\beta , \beta ) = t(0,1) = t \left( \varphi \begin{pmatrix} 0 & 0 \\ 0 &1 \end{pmatrix} \right),\) 
and so \(\varphi \)  is an isomorphism of algebras with trace between \(D_2^{t_{\alpha , \beta }}\)  and \(F_\alpha \oplus F_\beta \) . In the same way, if \(\alpha = \beta \) , we get a trace isomorphism between \(D_2^{t_{\alpha ,\alpha }}\)  and \(F_\alpha \oplus F_\alpha \) .
Hence, since \(D_2^{t_{\alpha , \beta }}, D_2^{t_{\gamma ,\gamma }}\) , \( D_2^{t_{\delta ,0}} \notin {tr}(A)\) , it follows that
\(A = B + J\) 
where \(B \cong F\)  and for all \(a = b+j \in A\) , \(tr(a) = tr(b+j) = \alpha b\) , with \(\alpha \ne 0\) .
In order to complete the proof we need to show that \(B + J\)  has polynomially bounded trace codimensions.
Notice that the following polynomials are trace identities of \(B + J\) :

\(\alpha \mbox{Tr}(x_1 x_2) - \mbox{Tr}(x_1) \mbox{Tr}(x_2) \equiv 0\) ,


\( \left( \mbox{Tr}(x_1) - \alpha x_1 \right)\cdots \left( \mbox{Tr}(x_{q+1}) - \alpha x_{q+1} \right)\equiv 0\) , where \(J^q \ne 0\)  and \(J^{q+1} = 0\) .

Modulo the first trace identity, every multilinear trace polynomial in the variables \(x_1\) , ..., \(x_n\)  is a linear combination of expressions of the type
\(\mbox{Tr}(x_{i_1}) \cdots \mbox{Tr}(x_{i_a}) x_{j_1} \cdots x_{j_b}\) 
where \( \left\lbrace  i_1, \ldots , i_a, j_1, \ldots , j_b \right\rbrace  = \left\lbrace  1, \ldots , n \right\rbrace  \)  and \(i_1 < \cdots < i_a \) . The second identity implies that we can suppose \(a\le q\) . Indeed if \(a\ge q+1\)  then one can represent the product of \(q+1\)  traces as a linear combination of elements with fewer traces by expanding the second trace identity. Therefore we may suppose \(a\le q\) .
We can further reduce the form of the latter polynomials. As we consider algebras with unit we can rewrite each monomial \(x_{j_1} \cdots x_{j_b}\)  as a linear combination of elements of the type
\(x_{p_1} \cdots x_{p_c} \left[ x_{l_1}, \ldots , x_{l_{k}} \right] \cdots \left[ x_{m_1}, \ldots , x_{m_{h}} \right].\) 
Here the commutators that appear are left normed, that is \([u,v] = uv-vu\) , \([u,v,w] = [[u,v],w]\)  and so on. Moreover
\(\left\lbrace  i_1, \ldots , i_a, p_1, \ldots , p_c, l_1, \ldots , l_k, \ldots , m_1, \ldots , m_h \right\rbrace  = \left\lbrace  1, \ldots , n \right\rbrace .\) 
By applying the Poincaré–Birkhoff–Witt theorem, we can assume further that \(p_1 < \cdots < p_c \) , and that the commutators are ordered. Therefore we can write each trace polynomial as a linear combination of elements of the type
\(\mbox{Tr}(x_{i_1}) \cdots \mbox{Tr}(x_{i_a}) x_{p_1} \cdots x_{p_c} [ x_{l_1}, \ldots , x_{l_{k_1}} ] \cdots [ x_{m_1}, \ldots , x_{m_{k_h}} ],\) 
where \( \left\lbrace  i_1, \ldots , i_a, p_1, \ldots , p_c, l_1, \ldots , l_{k_1}, \ldots , m_1, \ldots , m_{k_h} \right\rbrace  = \left\lbrace  1, \ldots , n \right\rbrace  \)  and \(i_1 < \cdots < i_a \) , \(p_1 < \cdots < p_c \) , \(a \le q\) . The algebra \(B\)  in the decomposition \(A=B+J\)  is commutative. Hence each commutator vanishes when evaluated on elements of \(B\)  only. Thus in order to get a non-zero element we must substitute in a commutator at least one element from \(J\) . Since \(J^{q+1} = 0\) , a product of \(q+1\)  commutators is a trace identity. Therefore we also get the restriction
\( K := k_1 + \cdots + k_h \le q\) . In this way
\(c_n^{tr}(A) &\le & \sum _{a=0}^q \binom{n}{a} \left( \sum _{K = k_1 + \cdots + k_h = 0}^q \binom{n-a}{K} \binom{K}{k_1, \ldots , k_h } k_1! \cdots k_h! \right) \\&=&\sum _{a=0}^q \dfrac{n(n-1) \cdots (n-a+1)}{a!}\left( \sum _{K = k_1 + \cdots + k_h=0}^q \dfrac{(n-a)!}{(n-a-K)!} \right) \\& \approx & cn^{2q}\) 
where \(c\)  is a constant. Hence \(A = B + J\)  has polynomial growth and the proof is complete.
